Problem 8 - Precalculus

Question

Use the pair of functions \(f\) and \(g\) to find the following values if they exist. $$ \begin{array}{lll} \bullet(f+g)(2) & \bullet(f-g)(-1) & \bullet(g-f)(1) \\ \bullet(f g)\left(\frac{1}{2}\right) & \bullet\left(\frac{f}{g}\right)(0) & \bullet\left(\frac{g}{f}\right)(-2) \end{array} $$ $$ f(x)=x^{2} \text { and } g(x)=\frac{3}{2 x-3} $$

Step-by-Step Solution

Answer

\((f+g)(2) = 7\), \((f-g)(-1) = \frac{8}{5}\), \((g-f)(1) = -4\), \((fg)(\frac{1}{2}) = -\frac{3}{8}\), \((\frac{f}{g})(0) = 0\), \((\frac{g}{f})(-2) = -\frac{3}{28}\).

Step 1: Calculate \((f+g)(x)\)

To find \((f+g)(2)\), use \(f(x) = x^2\) and \(g(x) = \frac{3}{2x-3}\). Calculate each function separately: \(f(2) = 2^2 = 4\) and \(g(2) = \frac{3}{2(2) - 3} = \frac{3}{4 - 3} = 3\). Now, add the results: \((f+g)(2) = 4 + 3 = 7\).

Step 2: Calculate \((f-g)(x)\)

For \((f-g)(-1)\), find \(f(-1)\) and \(g(-1)\). \(f(-1) = (-1)^2 = 1\) and \(g(-1) = \frac{3}{2(-1) - 3} = \frac{3}{-5} = -\frac{3}{5}\). Now subtract: \((f-g)(-1) = 1 - \left(-\frac{3}{5}\right) = 1 + \frac{3}{5} = \frac{8}{5}\).

Step 3: Calculate \((g-f)(x)\)

To find \((g-f)(1)\), compute \(g(1)\) and \(f(1)\). \(f(1) = 1^2 = 1\) and \(g(1) = \frac{3}{2(1) - 3} = \frac{3}{-1} = -3\). Now, subtract: \((g-f)(1) = -3 - 1 = -4\).

Step 4: Calculate \((f \cdot g)(x)\)

Find \((f g)\left(\frac{1}{2}\right)\) by evaluating \(f\left(\frac{1}{2}\right)\) and \(g\left(\frac{1}{2}\right)\). \(f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\) and \(g\left(\frac{1}{2}\right) = \frac{3}{2\left(\frac{1}{2}\right) - 3} = \frac{3}{1 - 3} = \frac{3}{-2} = -\frac{3}{2}\). Multiply the results: \((f g)\left(\frac{1}{2}\right) = \frac{1}{4} \cdot -\frac{3}{2} = -\frac{3}{8}\).

Step 5: Calculate \(\left(\frac{f}{g}\right)(x)\)

For \(\left(\frac{f}{g}\right)(0)\), first check \(g(0)\). \(g(0) = \frac{3}{2(0) - 3} = \frac{3}{-3} = -1\). Since \(g(0) eq 0\), find \(f(0) = 0^2 = 0\). Therefore, \(\left(\frac{f}{g}\right)(0) = \frac{0}{-1} = 0\).

Step 6: Calculate \(\left(\frac{g}{f}\right)(x)\)

Find \(\left(\frac{g}{f}\right)(-2)\) by first confirming \(f(-2)\) is not zero. \(f(-2) = (-2)^2 = 4\). Since \(f(-2) eq 0\), compute \(g(-2) = \frac{3}{2(-2) - 3} = \frac{3}{-7}\). Thus, \(\left(\frac{g}{f}\right)(-2) = \frac{-\frac{3}{7}}{4} = -\frac{3}{28}\).

Key Concepts

Function Addition

Function addition involves combining two functions by summing their outputs for a given input. It means that for two functions, \(f(x)\) and \(g(x)\), the new function \((f+g)(x)\) equals \(f(x) + g(x)\).

• Start by calculating each function value separately at the input of interest.
• Once you have these values, you simply add them together.

For example, if \(f(x) = x^2\) and \(g(x) = \frac{3}{2x-3}\), and you need to find \((f+g)(2)\), you first compute \(f(2) = 4\) and \(g(2) = 3\).
Thus, \((f+g)(2) = 4 + 3 = 7\).
This is a straightforward process to blend functions by adding their outputs at a specific point.

Function Subtraction

Function subtraction works similarly to addition but involves taking the difference of the two functions. For two functions \(f(x)\) and \(g(x)\), the function \((f-g)(x)\) is defined as \(f(x) - g(x)\).
To perform this operation:

• Calculate the outputs of the functions at the desired input separately.
• Subtract the value of \(g(x)\) from \(f(x)\).

For example, with \(f(x) = x^2\) and \(g(x) = \frac{3}{2x-3}\), to find \((f-g)(-1)\), compute \(f(-1) = 1\) and \(g(-1) = -\frac{3}{5}\).
Then, \((f-g)(-1) = 1 - (-\frac{3}{5}) = \frac{8}{5}\).
This method helps us find the difference specifically dictated by each function's formula.

Function Multiplication

Multiplying functions is a bit more complex because it involves the product of the outputs. If you have two functions \(f(x)\) and \(g(x)\), then \((fg)(x)\) is defined as \(f(x) \cdot g(x)\).

• First, evaluate each function separately at the desired input.
• Multiply these two values together.

Using \(f(x) = x^2\) and \(g(x) = \frac{3}{2x-3}\), to compute \((fg)\left(\frac{1}{2}\right)\), we get \(f\left(\frac{1}{2}\right) = \frac{1}{4}\) and \(g\left(\frac{1}{2}\right) = -\frac{3}{2}\).
The product is \(-\frac{3}{8}\), showing how function multiplication blends the separate influences of the two functions.

Function Division

Function division requires extra caution, as it involves dividing the output of one function by another. For functions \(f(x)\) and \(g(x)\), \(\left(\frac{f}{g}\right)(x)\) means \(\frac{f(x)}{g(x)}\), provided \(g(x) eq 0\).

• First, evaluate both functions at the input.
• Ensure that the denominator, \(g(x)\), is not zero to avoid division by zero.
• Divide the result of \(f(x)\) by the result of \(g(x)\).

For example, evaluating \(\left(\frac{g}{f}\right)(-2)\), where \(f(x) = (-2)^2 = 4\) ensures the denominator is non-zero.
Then compute \(g(-2) = -\frac{3}{7}\). So, \(\left(\frac{g}{f}\right)(-2) = -\frac{3}{28}\).
This shows the delicate balance in function division where we cautiously navigate zeros in the denominator.